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stira [4]
2 years ago
5

Calculate the density (in grams per milliliter) of the following gas at 97 °C and 755 mm Hg Sulfur dioxide (SO2)

Chemistry
1 answer:
Diano4ka-milaya [45]2 years ago
3 0

Answer:

Density = 2.09×10⁻³ g/mL

Explanation:

We solve this problem from the Ideal Gases Law.

P . V = n . R . T

We know that in STP conditions 1 mol of any gas occupies 22.4 L. So we can compare both situations.

22.4 L = 22400 mL

(P₁ . V₁) / T₁ = (P₂ . V₂) / T₂

We make some conversions:

97°C + 273 = 370 K

755 mmHg . 1 atm / 760mmHg = 0.993 atm

We replace data: (1 atm . 22400 mL) / 273 K = (0.993 atm . V₂) / 370 K

((1 atm . 22400 mL) / 273 K) . 370K = 0.993 atm . V₂

V₂ = 30359 atm . mL / 0.993 atm → 30573 mL

Under those conditions (97 °C and 755 mm Hg) 1 mol of gas is contained in 30573 mL.

We determine the amount of gas: 1 mol of SO₂ weighs 64.06g

Density = m/V → 64.06 g / 30573 mL = 2.09×10⁻³ g/mL

You might be interested in
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
You have an unknown quantity of oxygen at a pressure of 2.2 atam, a volume of 21 liters and a temperature of 87 Celsius. How man
laila [671]

<span>Let's assume that the oxygen gas has ideal gas behavior. 
Then we can use ideal gas formula,
      PV = nRT</span>


Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

<span>
P = 2.2 atm = 222915 Pa
V = 21 L = 21 x 10</span>⁻³ m³

n = ?

R = 8.314 J mol⁻¹ K⁻¹

<span> T = 87 °C = 360 K

By substitution,
</span>222915 Pa x 21 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻<span>¹ x 360 K
                                       n = 1.56</span><span> mol</span>

<span>
Hence, 1.56 moles of the oxygen gas are </span><span>left for you to breath.</span><span>
</span>
6 0
3 years ago
Calculate the percentage composition of Mg3 (Po4)2
lutik1710 [3]
Mg3(PO4)2 - the molar mass would be 262g/mol, which is 100%

Atomic mass of Mg is 24, since we have 3Mg we multiply by 3 and get a mass of 72

262 : 100% = 72 : x%

x = 72*100 / 262

x = 27.5%

And do that for every element — get the molar mass of P and multiply by 2, use a ratio, and get the molar mass of O and multiply by 8 and use ratios :)
7 0
2 years ago
Identify the substances that are likely to dissociate in water. Check all that apply
almond37 [142]
The hydroxide give it away
8 0
2 years ago
Adding more baking soda and vinegar to the bag in the Air Bag Lab resulted in faster production of gaseous products True False
nika2105 [10]

Answer:

True

Explanation:

The more of each substanse that you add to the bag will cause it to produce faster and more gas.

5 0
2 years ago
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